Question

Find the angle(s) at which the curves $y^2=4xand{x}^2=4y$ intersect:

• A. $\theta =ta{n}^{-1}\left(\frac{1}{3}\right)$
• B. $\theta =ta{n}^{-1}\left(\frac{1}{4}\right)$
• C. $\theta =ta{n}^{-1}\left(\frac{3}{4}\right)$
• D. $\theta =ta{n}^{-1}\left(\frac{4}{3}\right)$

Answer 1

The correct option is C $\theta =ta{n}^{-1}\left(\frac{3}{4}\right)$

$y^2=4x$ $x^2=4y$

$y=2\sqrt{x}{x}^2=4\times 2\sqrt{x}$

$x^2=8\sqrt{x}$

Squarring on both sides

$x^4=64x$

$x=0x=4$

$y=0$

$y=4$

$\text{Tangent to}{y}^2=4x\text{at}(4,4)\text{is}$

$y\left(4\right)=2(x+4)$

$2y=x+4$

Tangent to $x^2=4y$ at (4,4) is

$4x=2(y+4)$

$2x=y+4$

Angle between the tangents is equal to angle between the curves

$\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$

$\tan \theta =\left|\frac{\frac{1}{2}-2}{1+1}\right|=\frac{3}{4}$

$\theta ={\tan }^{-1}\left(\frac{3}{4}\right)$